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Lógica matemática/ f(a,b,c)=(avb)⊕(c⇒¬a)

Expresión f(a,b,c)=(avb)⊕(c⇒¬a)

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    Solución

    Ha introducido [src] 
    (a∨b)⊕(c⇒(¬a))
    $$\left(a \vee b\right) ⊕ \left(c \Rightarrow \neg a\right)$$
    Solución detallada
    $$c \Rightarrow \neg a = \neg a \vee \neg c$$
    $$\left(a \vee b\right) ⊕ \left(c \Rightarrow \neg a\right) = \left(a \wedge c\right) \vee \left(\neg a \wedge \neg b\right)$$
    Simplificación [src]
    $$\left(a \wedge c\right) \vee \left(\neg a \wedge \neg b\right)$$ 
    (a∧c)∨((¬a)∧(¬b))
    Tabla de verdad 
    +---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1      |
+---+---+---+--------+
| 0 | 0 | 1 | 1      |
+---+---+---+--------+
| 0 | 1 | 0 | 0      |
+---+---+---+--------+
| 0 | 1 | 1 | 0      |
+---+---+---+--------+
| 1 | 0 | 0 | 0      |
+---+---+---+--------+
| 1 | 0 | 1 | 1      |
+---+---+---+--------+
| 1 | 1 | 0 | 0      |
+---+---+---+--------+
| 1 | 1 | 1 | 1      |
+---+---+---+--------+
    FND [src]
    Ya está reducido a FND
    $$\left(a \wedge c\right) \vee \left(\neg a \wedge \neg b\right)$$ 
    (a∧c)∨((¬a)∧(¬b))
    FNCD [src]
    $$\left(a \vee \neg b\right) \wedge \left(c \vee \neg a\right)$$ 
    (a∨(¬b))∧(c∨(¬a))
    FNDP [src]
    $$\left(a \wedge c\right) \vee \left(\neg a \wedge \neg b\right)$$ 
    (a∧c)∨((¬a)∧(¬b))
    FNC [src]
    $$\left(a \vee \neg a\right) \wedge \left(a \vee \neg b\right) \wedge \left(c \vee \neg a\right) \wedge \left(c \vee \neg b\right)$$ 
    (a∨(¬a))∧(a∨(¬b))∧(c∨(¬a))∧(c∨(¬b))
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